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Erdős problem / erdos

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Problem 1

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  1. vf_7746daa976a29829

    theoretical

    Erdős Problem #1: declared status 'open'. Formalized: yes. If A{1,,N}A\subseteq \{1,\ldots,N\} with A=n\lvert A\rvert=n is such that the subset sums aSa\sum_{a\in S}a are distinct for all SAS\subseteq A thenN2n.N \gg 2^{n}. Current best: The trivial lower bound is N2n/nN \gg 2^{n}/n, since all 2n2^n distinct subset sums must lie in [0,Nn)[0,Nn). Erd\H{o}s and Moser [Er56] provedN(14o(1))2nn. N\geq (\tfrac{1}{4}-o(1))\frac{2^n}{\sqrt{n}}.(In [Er85c] Erd\H{o}s offered \100foranyimprovementoftheconstant100 for any improvement of the constant 1/4here.)Anumberofimprovementsoftheconstanthavebeengiven(see\citeSt23forahistory),withthecurrentrecord here.) A number of improvements of the constant have been given (see \cite{St23} for a history), with the current record \sqrt{2/\pi}firstprovedinunpublishedworkofElkiesandGleason.TwoproofsachievingthisconstantareprovidedbyDubroff,Fox,andXu\citeDFX21,whoinfactprovetheexactbound first proved in unpublished work of Elkies and Gleason. Two proofs achieving this constant are provided by Dubroff, Fox, and Xu \cite{DFX21}, who in fact prove the exact bound N\geq \binom{n}{\lfloor n/2\rfloor}.Prize:. Prize: 500. OEIS: A276661. Tags: additive combinatorics, number theory.

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